सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

261
Fig. 44
Verse 41. Alternate method of obtaining K
101530/H sin λ = para (say) where λ is the Sāyana
longitude of the Sun ; then, (Para × k) / s = K where K is
un-mandala-Karṇa
First half of Verse 42. To obtain K when the Sun is
on the prime-vertical— Para × s/k = Samavṛttakarṇa.
Comm. From fig. 19, from the similarity of triangles
BD ☉ and CMA, B ☉ / CA = ☉ D / AM ie. ☉ D = (B ☉ × AM) / CA ie.
H sin δ = (H sin λ H sin ω) / R (a)
Then consider the similarity of the first and the sixth
latitudinal triangles ; then Unmandala Sanku / Krāntijyā = s / k (b)
where s is the equinoctial shadow and k the Viṣuvat-
Karṇa. Again taking that ☉ the Sun lies on the unman-

262
dala in figure 44, (H cos z) / 12 = R / K = (Unmandala Śaṅku) / 12 (c)
Eliminating Krāntijyā and Unmandala Śaṅku from
(a), (b) and (c) (Unmandala Śaṅku) / (H sin λ H nis ω/R) = s / k
∴ (12R / K) / (H sin λ H sin ω / R) = s / k ie. 12R² / (KH sin λ H sin ω) = s / k
∴ K = (12R² × k) / (s H sin λ H sin ω). Here 12R² / (H sin ω) = (12 × 3438²) / 1397
= 101531 ; but Bhāskara has taken 101530 taking a
more correct value of R. Then 101530 / (H sin λ) is symbolized as
para so that para × k / s = K = Unmandala Karṇa. Regard-
ing the Samavṛttakarṇa, in the place of (b) above we have
(Sama-Śaṅku) / Krāntijyā = k / s (b') by the similarity between the
first and the fifth latitudinal triangles. Equation (c)
holds good with respect to any H cos z and the corres-
ponding K since 12 R = K × Śaṅku and 12 R is a cons-
tant. Noting therefore R / K' = (Sama-Śaṅku) / 12 (c')
eliminating Krāntijyā and Sama-Śaṅku among (a), (b'),
(c'), we shall have
K = (12 R² s) / (k H sin λ H sin ω) = para × s / k as stated.
Second half of Verse 42. To obtain the Dinārdha-
karṇa from the Unmandalakarṇa.
(Un-mandalakarṇa × Charajyā) / Antyā = Dinārdhakarṇa.
Comm. We have equation (c) above stating
12 R = K × Śaṅku. (c) But

263
Iṣṭa Śaṅku / Iṣṭa Hṛti = cos φ = constant = Dinārdha Śaṅku / Hṛti
= Sama-Śaṅku / Taddhṛti = Unmaṇḍala Śaṅku / Kujyā (d) (23)
Again by virtue of the proportionality of
Iṣṭa Hṛti / Iṣṭāntyā = Hṛti / Antyā = Kujyā / Charajyā (e) (24)
We have Ishta-Śaṅku / Ishtāntyā = Dinārdha Śaṅku / Antyā
= Unmaṇḍala Śaṅku / Charajyā
∴ Iṣṭa Karṇa × Iṣṭāntyā = Dinārdha Karṇa × Antyā
= Unmaṇḍala Karṇa × Charajyā (f) (25)
∴ Dinārdha Karṇa = (Unmaṇḍala Karṇa × Charajyā) / Antyā
as stated in the verse.
Verse 43. (Unmaṇḍala Karṇa × Kṣitijyā) / Hṛti
= (Sama Vṛtta Karṇa × Taddhṛti) / Hṛti = Dinārdha Karṇa.
Comm. From (c) and (d) above Dinārdha Karṇa ×
Hṛti = Sama Karṇa × Taddhṛti = Unmaṇḍala Karṇa ×
Kujyā (g). Khitijyā is the same as Kujyā.
From this the statement follows :
Verse 44. The ancient Achāryas found the gnomonic
shadows when the Sun is on the meridian, prime-vertical
and the Kona-Vṛtta (ie, Vertical when the northern or
southern Hindu azimuths are 45°) by different methods.
I consider him to be the very Sun illuminating the lotus-
faces of aitronomers, if anybody could give a method to
find the shadow in any required direction, which holds
good in all cases universally.

264
Comm. Evident.
Verse 45. Definition of Dikjyā H sin (azimuth). The
angle between any vertical and the Prime-Vertical
measured on the horizon is what is called Digamsa and its
H́ sine is known as Dik-jyā either in the Eastern hemi-
sphere or the Western.
Comm. In modern astronomy azimuth is measured
along the horizon from the north point towards the east
point round the horizon. In Hindu Astronomy however,
the azimuth is measured from the East point on either
side and from the West point also on either side specifying
whether it is north or south.
Verse 46 and first half of 47. To obtain the
gnomonic shadow in any arbitrary direction.
Assume Rs / (H sin a) as the equinoctial shadow and
obtain the H sine of the corresponding latitude L. Then
the product of that H sin L and H sin δ divided by
H sin ϕ will give H sin D where D is a hypothetical decli-
nation. With the new L and this D, as the hypothetical
latitude and declination, obtain the meridian zenith
distance by the formula Z + D = ϕ, and through this
m.z d. obtain the shadow, which will be the shadow in the
required direction namely 12 tan (ϕ ± D).
Comm. Let gL be the gnomonic shadow on the
equinoctial day in a given direction given by a°
Digamsa (the Hindu azimuth) and let gN be the shadow in
the same direction on any day. (fig. 45) We know that the
extremity of the gnomonic shadow on the equinoctial day
traces a straight line parallel to the East-West line Eω
at a distance of the equinoctial shadow s because the
Equatorial plane passing through the foot of the gnomon

265
Fig. 45
and that passing through the top of the gnomon being
parallel planes cut the horizontal plane in parallel straight
lines. (This will be also proved analytically subsequently).
Hence LM = s. Now from the figure
LM / gL = AB / gA = H sin a / R ∴ gL = H sin a / Rs I
This gL is spoken of as Iṣṭa-Drikmandala palabhā
because it is the shadow on the equinoctial day in
any vertical. LN is the increment in the shadow on
account of declination and we have to compute this and
correlate gL and LN. For this refer to figs. 46 and 47.
In fig. 46, QRT is the equator, so that when the Sun is on
the equator on the equinoctial day in the direction given
by ZS, ZT is the zenith-distance. Let ZS be the zenith-
distance of the Sun in the same direction on any day
From the analogy of finding H sin δ from H sin λ, from
this figure
34

266
H sin SR = (H sin ST × H sin T̂) / R II and
H sin φ = (H sin ZT × H sin T̂) / R III so that
(H sin SR) / (H sin φ) = (H sin ST) / (H sin ZT) ∴ H sin ST = (H sin SR / H sin φ) × H sin ZT.
Noting that SR = δ and putting ST = D
H sin D = (H sin δ / H sin ϕ) × H sin ZT.
[Diagrams: Fig. 46 and Fig. 47 showing spherical triangles with vertices P, Q, N, R, S, T, D, δ]
Fig. 46 Fig. 47
The same formulae are derivable from fig. 47 also; only
in fig. 46 while there is a decrement in the shadow of the
day as compared with the shadow on the equinoctial day,
in fig. 47, there is an increment. This is seen from the
decrease and increase of ST in the zenith-distance ZT of
the equinoctial day in the given direction. Now corre-
lating fig. 45 with figures 46 and 47, the shadow gL
pertains to the zenith-distance ZT on the equinoctial day
whereas the shadows gN pertains to the zenith-distance on
the day concerned in the same direction. We have,
S / √(12² + S²) = (H sin z) / R so that RS / √(12² + S²) = H sin z where
S is the shadow at any instant when the zenith-distance
is z. The process indicated by saying ‘Obtain H sin φ

267
construing Rs / (H sin a) as the equinoctial shadow', means
computing H sin ZT and there from ZT from the shadow
gL of fig. 45. Then the process indicated by saying
“Obtain H sin D = (H sin ZT × H sin δ) / (H sin φ) and therefrom
D'' means computing ST.
Then clearly ZS = ZT ± ST, ie. the required zenith-
distance is got by what is technically called Samskāra
between ZT and ST as is stipulated between φ and δ to
obtain Z from the formula Z ± δ = φ (The word
Samskāra was defined as meaning addition when the
directions are the same and difference when they are
opposite). Then the gnomonic shadow is got from this
zenith-distance using the formula S = (12 H sin z) / (H cos z) .
Thus the procedure adopted by Bhāskara was con-
ceived by him first having Fig. 45 before him and then
using figures 46 and 47. In this particular process,
H sin a is given and H sin δ also, which means that it
is sought to find the shadow on a given day in a given
direction.
Incidentally we shall find the locus of the extremity
of the gnomonic shadow during the course of a day. Let
in fig. 48 g represent the gnomon's foot, and S the shadow
whose extremity is p. Required to find the locus of p.
Take the gnomon to be of unit length so that the length of
the shadow S = 12 tan z becomes tan z here. Take Eω
and sn the east-west line and the north-south as the
axes. Then we have
x² + y² = tan² z (1). But we have from the triangle
PZS sin δ = sin φ cos z + cos φ sin z sin a
ie. sin δ / (cos φ cos z) = tan φ + tan z sin a = tan φ + y (2)
ie. sec z / A = y + tan φ when A = cos φ / sin δ .

268
Fig. 48
But sec z = √(1 + tan² z) = √(x² + y² + 1)
∴ √(x² + y² + 1) / A = y + tan Φ which reduces to
x² + y² (1 - A²) - 2A² y tan Φ + 1 - A² tan² Φ = 0 (3)
From this it is evident that the locus is an ellipse or
parabola or hyperbola according as A ⋚ 1; also it will
be seen that the eccentricity is A. The locus is wrongly
stated to be always a hyperbola in some text books. For
it to be an ellipse A < 1 ie. cos Φ < sin δ ie. δ > 90 - Φ
ie. Φ + δ > 90. In such latitudes and under such decli-
nations, it will be an ellipse ie. at a place just north of
the place where the perpetual day just begins the locus will
be an ellipse. Hence in the arctic region it will be always
an ellipse; and in the place just at which the perpetual
day begins it will be a parabola and in the lower latitudes
it will be a hyperbola, ie. it will be a parabola where the
latitude Φ is given by 90 - δ.
When Φ = 90°, A = cos Φ / sin δ = 0 provided δ ≠ 0. If,
however, in addition δ = 0, A becomes indeterminate, but
we may note then, that the Sun will be circling round the
horizon on that equinoctial day at the north pole. We

269
may further note that at the north pole, the altitude of
the Sun is always δ so that the length of the shadow cast is
always equal to cot δ and this will be infinite when δ = 0.
If now φ = 90°, and δ ≠ 0, though A = cos φ / sin δ be
comes zero, A tan φ will not be zero because A tan φ =
cos φ / sin δ × tan φ = sin φ / sin δ = 1 / sin δ (∵ φ = 90°). On the
other hand A² tan φ = A × A tan φ = / sin δ = 0 because
A = 0. Thus the term containing y in eqn. (3) vanishes.
∴ The equation reduces to
x² + y² = A² tan² φ - 1 = cosec² δ - 1 = cot² δ (ie. +ve)
ie. at the north pole the locus will be a circle with radius
cot δ. When
A = ∞ the locus √(x² + y² + 1) / A = y + tan φ becomes
y = - tan φ which means that the hyperbola degenerates
into the straight line which is parallel to the east-west
line and is in the north at a distance of tan φ ie. s, the
equinoctial shadow since the length of the gnomon
is taken to be unity. In particular when A tan φ = 1 ie.
φ = δ, the constant in (3) is zero, so that the locus
passes through the foot of the gnomon as is also evident
from the fact that the Sun passes through the zenith.
Taking a northern latitude say 17°, the loci of the
extremity of the shadow are shown in fig. 48A (page 270)
on important days when δ=ω, when δ=0, when δ = - ω,
when δ = φ. The maximum mid-day shadow is
tan (φ + ω), when δ = - ω taking the gnomon's length
to be unity; this shadow is cast north of the gnomon along
the south-north line through the gnomon, on Dec. 23rd of
the year. The minimum length of the mid-day shadow
occurs when δ = φ, the shadow being zero and being at the
foot of the gnomon, the Sun being then just overhead. The
maximum shadow cast south of the gnomon at mid-day
is tan (ω - φ).

270
Shadow on Dec. 23
No Shadow in this direction on Dec 23
Immediately before 21st March
B C A On 21st March
Immediately after 21st March
No Shadow in this direction on 22nd June
W B C A E
JUNE when δ=φ
On 22nd JUNE
Fig. 48A Showing the locus of the extremity of the shadow on different
days at a latitude of 17°
Note. OA, OB are the shadows computed by Bhāskara when the Sun is on
the prime-vertical.
In the method of finding the shadow under verse 46,
we perceive Bhāskara's genius in (1) looking upon the
shadow as being made up of two segments namely that due
to φ and that due to the declination (2) in conceiving
what he calls Iṣṭa-drik-mandala palabhā, and Iṣṭa-drik-
mandala Krānti and (3) in deriving the equation
H sin D = (H sin ZT × H sin δ) / (H sin φ)
Latter half of verse 47 and verse 48. Something
to be noted.
In computing the shadow in a given direction, there
may be two shadows at times in the northern hemi-
sphere. When H sin a < Agrā and there will be none in
the southern. To compute the second shadow we have to
take 180 − L also as the latitude where L is the latitude
computed, and proceed in the same way as we have done
before.

271
Fig. 49
Comm. This too exhibits Bhāskara's genius. (Ref.
fig. 49). Let MQR₂ be the equator whose pole is p. Let
T₁ S₁ Z S₂ T₂ be the circle of azimuth a (Hindu azimuth).
Let SS₁ S₂ be the diurnal circle of the Sun cutting the
above circle of azimuth at S₁ and S₂, so that ZS₁ and ZS₂
are the two solutions giving the two zenith-distances
which give two shadows in the given direction. H sin MS
= Agrā ; evidently MẐS > MẐS₁ ie. H sin a < Agrā as
stipulated. ZT₁ and ZT₂ give the zenith-distances in the
given direction when the Sun is on the equator. S₁ T₁
and S₂ T₂ are the decrements in the zenith-distances on
account of declination δ (= S₂ R₂ or S₁ R₁). If MẐS₁ were
greater than MẐS ie. if H sin a > Agrā, we would have
lost the position S₁ ie. we would have had only one shadow

272
in the afternoon in the given direction and no shadow in
the morning.
Analytically, the event of having two shadows arises
on account of the following circumstance. When we are
asked to find Iṣṭākṣajyā from Rs / (H sin a) the shadow in the
given direction on the equinoctial day (Ref. gL fig. 45)
the L for this given value of the shadow is given by
H sin L = RS / √(12² + S²) where S = Rs / (H sin a). We know
sin θ = a has two solutions, θ₁ and 180 − θ₁. Hence L
will have two values L₁ and 180 − L₁. So, Bhāskara has
asked us to compute D₁ and D₂ from L₁ and L₂ and thus
have the two solutions.
Fig. 50

273
Note (1) When Bhāskara said 'If H sin a < Agrā'
he had in mind evidently the azimuth circle MZN which
cuts the diurnal path A Q₁ R₁ at S₁ and S₂. At S₁ the
azimuth EZM < EZA so that he stipulated that H sin a
should be less than the Agrā. But, let the diurnal path of
the Sun be Q₂ R₂ where Q₂ falls in between z and p. In
such a case we know that the azimuth does not take all
values but has a maximum where the vertical touches the
diurnal path at T. From PTZ where T is a right angle,
taking PZT = 90 - a, a being the Hindu azimuth we have
by Napier’s rule sin PT = sin ZP sin (90 - a) or
sin (90 - δ) = sin (90 - φ) sin 90 - a or cos δ = cos φ cos a.
If a has a lesser value than is given by this equation,
the diurnal path does not cut the azimuth circle
ie. if cos a > cos δ / cos φ, there will be no shadow in the given
direction even though the situation satisfies Bhāskara’s
condition namely H sin a should be less than Agrā.
Bhāskara has overlooked this case. This may be seen
analytically also as follows. We have from the spherical
triangle PZS, sin δ = sin φ cos z + cos φ sin z sin a
= A cos z + B sin z (say). We know, the maximum
value of A cos z + B sin z is √(A² + B²) which is here
√(sin² φ + cos² φ sin² a) = √(sin² φ + cos² φ - cos² φ cos² a)
= √(1 - cos² φ cos² a). Thus there will be no solution for
z if the quantity on the left hand side namely
sin δ > the above max. value ie. if
sin δ > √(1 - cos² φ cos² a) ie. if sin² δ > 1 - cos² φ cos² a
ie. if cos² δ > cos² φ cos² a ie. if cos δ > cos φ cos a
ie. if cos a > cos δ/cos p as derived above.
Hence even if H sin a > Agrā, there need not be a
shadow at all in the given direction. In other words when
the Hindu azimuth given is very small and when the decli-
nation is too great north or south, there may not be a
35

274
shadow in the given direction. Bhāskara gives an example
where he gets two shadows on a day taking the moments
when the Sun is on the prime-vertical. In fact having
this case of the East-West shadows alone, he conceived
that two shadows could be had in a given direction under
particular conditions.
He chooses a place of s = 5″ ie. a place of latitude
22° – 37′ (Bhāskara often gives this latitude which night
indicate that he was probably residing in that latitude
which passes through approximately Itarsi). He takes a
day when the Sun’s declination is given by H sin δ = 780
ie. δ = 13° – 7′. Then the Sama-Śanku is given by
R sin δ
───────── (comparing the second and the fifth latitudinal
sin φ
triangles
Sama-Śanku Krantijyā H sin δ
──────────── = ─────────── = ─────────
R H sin φ H sin φ
RH sin δ R sin δ
∴ Sama-Śanku = ────────── = ─────────)
H sin φ sin φ
3438 × 780
= ────────── = 2028 approximately. I
1322 – 18
R sin δ
Also Agrā = ───────── = 845.
cos φ
Knowing the Sama-Śanku, the East-West shadow may
be taken to be determined. Now Bhāskara proceeds to show
that at the time of having the second shadow also, in the
same East-West direction, the Śanku will be the same
Sama-Śanku itself. For this, proceeding according to the
Rs
method indicated in the verse, “Taking ─────── to be the
H sin a
equinoctial shadow etc.” we have
Rs 3438 × 5
─────── = ──────────── = what is called Kha-hara Rāsi.
H sin a 0

275
Taking this to be the equinoctial shadow
H sin L = Rs / √(12² + s²) = R itself (Dealing this way
with Kha-hara Rās'is is prohibited in modern mathematics
but Bhāskara adds at the end of the commentary that
dealing with them cautiously does not effect computations
which is of course true, for when the equinoctial shadow is
infinity φ = 90° so that H sin φ = R as got). Hence
L = 90° and 180° - 90° = 90° = L'. Then H sin D =
(R × 780) / (1322 - 18) = same as Sama-S'anku obtained in I = H cos z
so that D = 90 - z. Now from the equation z + δ = φ,
z = φ - δ = L' - D = 90° - (90 - z) where z is the
zenith-distance when the Sun is on the prime-vertical.
∴ The zenith-distance is again the same z. In other
words, the second zenith-distance is also that when the
Sun is on the prime-vertical. Bhāskara has given this
example just to obtain the second shadow as well and he
has chosen the event of the Sun being on the prime-
vertical to show that the procedure indicated by him may
be verified to hold good.
Verses 49, 50. Alternate method to find the shadow.
Let R² s² + H sin² a × 12² = prathama where s =
equinoctial shadow and a the Hindu azimuth. Let Anya
= RsA where A is the Agrā. Divide the prathama and
Anya by (H sin² a - A²) and still call them prathama and
Anya. Then
K = √(Adya + Anya²) ± Anya where K is the Chayā-
Karṇa.
Comm. Let K be the required Chayā-Karṇa.
Then Karnāgrā = KA / R = s + b where b is the bhuja.

276
∴ b = KA / R - s. But (H sin a) / b = R / S where S is the
shadow so that
S = bR / (H sin a) = (KA / R - s) R / (H sin a) = √(K² - 12²)
∴ (KA - sR) = H sin a √(K² - 12²)
ie. K² A² + s² R² - 2 AsRK = H sin² a (K² - 12²)
∴ K² (A² - H sin² a) - 2 AsRK = - s² R² - 12² H sin² a
∴ K² (H sin² a - A²) + 2 AsRK = 12² H sin² a + s² R² I
This quadratic is of the form ax² + 2bx = c
ie. x² + 2 (b/a) x = c/a II Here 'c'/a is called Adya and b/a Anya.
The solution of the above equation is given by x = - b/a ±
√(b²/a² + c/a)
ie. - Anya ± √(Anya² + Adya). III
When b = KA / R + s, putting - s in the place of s in I,
K = Anya ± √(Anya² + Adya) IV
Out of the four solutions given by III and IV we have
taking the positive solutions
K = √(Anya² + Adya) ± Anya as stated.
Verse 51. If H sin a < A, then in the northern
hemisphere ie. where δ is north,
± √(Anya² - Adya) + Anya = K.
Comm. We have initially put Anya = H sin² a - A².
If H sin a < A, them to avoid a negative value for the

277
Anya, we could put Anya = A² - H sin² a. As a matter
of fact in verse 50, we are asked to take H sin² ~ A²,
as Bhāskara wanted that the second case also be included.
Thus putting Anya = A² - H sin² a, equation I becomes
K² (A² - H sin² a) - 2 AsRK = - (Adya) so that
K = Anya ± √(Anya² - Adya) as given.
Verse 52. The Bhuja is to be obtained through
Karṇāgra from the equation a = b + s (already proved)
and Rb / S = H sin a {ie. S = bR / H sin a as already proved}.
This H sin a will be the same in the case of obtaining
two values of K ie. two shadows one in the morning and
the other in the afternoon, (the only difference being that
they will be on alternate sides of the East-West line).
Verses 53 and 54. Obtaining the shadow when time
is given.
In the two previous examples the magnitude of the
shadow was obtained in a given direction ; now we shall
obtain the same when time is given. The word unnata
stands for the time that has elapsed after Sun-rise or that
which is the balance of the day time. The unnata sub-
tracted from half-the-day gives what is called Nata.
The H sine of the unnata minus Chara or increased
by the Chara according as the Sun's declination is north
or south, is called Sūtra ; this multiplied by the H cos δ
and divided by the radius, is called Kalā.
Comm. (Ref fig. 51) The time measured by the arc
MN, that is the time in between the moment when the Sun
S is on the horizon and the moment when he is at L is
called the unnata ie. the time measured after rising and
the time measured by the arc NQ ie. the time in between

278
Fig. 51
the moments when the Sun is at L and when he reaches
the meridian. is called Nata. In modern astronomy this
Nata is known as the hour angle h and unnata = H - h
where H is the rising hour angle. The time measured by
the arc ME is called Chara. Thus unnata-chara = EN
and H sin EN H sin (90 - h) = H cos h is called Sutra
which is BN shown in fig. 52 representing the Equator.
The corresponding line bn in fig. 53 which represents the
diurnal circle, is known as Kalā. Thus Sūtra = H cos h (26)
and Kalā = (H cos h × H cos δ) / R (27).
When the Sun is in the Southern hemisphere, unnata
is measured by the arc SL or MN ie. (H - h) where H is

279
Fig. 52
Fig. 53

280
the rising hour angle and ℎ = L P̂ Q and Chara by the arc
EM and Sūtra = H sin (MN + EM) = H sin (unnata +
chara) = H sin EN = H sin (90 - ℎ) = H cos ℎ.
Verse 55. Sūtra multiplied by Kujyā and divided by
Charajyā will be also Kalā and Kalā multiplied by the Koti
and divided by the Karṇa of any latitudinal triangle will
be Iṣṭa yaṣṭi.
Comm. In as much as Kalā is the corresponding line
in the diurnal circle to the Sūtra in the plane of the
celestial equator
Sūtra / Kalā = R / (H cos δ) = Charajyā / Kujyā ∴ (Sūtra × Kujyā) / Charajyā = Kalā
In verse 33, we saw that
Dinārdha Sanku — Unmandala Sanku = Yaṣṭi.
This Iṣṭa-yaṣṭi will be therefore the perpendicular dropped
from 𝑛, the Sun's position in the diurnal circle on the
plane parallel to the horizon and passing through the
head of the Unmandala Sanku ie. passing through B and
parallel to the horizon in fig. 21. Since the angle between
the diurnal plane and the vertical plane of the yaṣṭi is
equal to φ the latitude, the Iṣṭayaṣṭi forms a latitudinal
triangle with the Kalā, it being the Koti or side opposite
to the angle 90 - φ and the Kalā being the hypotenuse,
∴ Yaṣṭi / Kalā = cos φ = Koti of a latitudinal triangle / Karṇa of a latitudinal triangle
∴ Iṣṭayaṣṭi = Kalā × (Koti of a latitudinal triangle) / Karṇa
When the Sun is on the meridian, Iṣṭayaṣṭi becomes yaṣṭi
of verse 33.
The formula for Iṣṭayaṣṭi is therefore
(Kalā × H cos φ) / R = (H cos δ · H cos ℎ) / R × (H cos φ) / R
from formula (27)
= (H cos φ · H cos δ · H cos ℎ) / R² 27'